检查给定的解码字符串是否能被 6 整除
给定由小写字符组成的字符串 str ，任务是根据给定规则进行更改后，检查该字符串是否可被 6 整除:
- ‘a’改为 1 。
- ‘b’改为 2 …
- 同样地，‘z’变成了 26 。
例如，字符串“abz”将更改为 1226 。
示例:
输入:str = " ab " T3】输出:是 “ab”相当于 12，可被 6 整除。
输入:str = " ABC " T3】输出:否 123 不能被 6 整除。
方法:只要一个数的所有位数之和能被 3 整除，且该数的最后一个位数能被 2 整除，就可以用一个简单的数学技巧解决这个问题。求形成数的位数之和，存入变量 sum 。此外，找到号码的最后一位数字并存储在最后一位数字中。 现在，如果总和被 3 整除，最后一位数字被 2 整除，则打印“是”否则打印“否”。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the sum
// of the digits of n
int sumDigits(int n)
{
int sum = 0;
while (n > 0) {
int digit = n % 10;
sum += digit;
n /= 10;
}
return sum;
}
// Function that return true if the
// decoded string is divisible by 6
bool isDivBySix(string str, int n)
{
// To store the sum of the digits
int sum = 0;
// For each character, get the
// sum of the digits
for (int i = 0; i < n; i++) {
sum += (int)(str[i] - 'a' + 1);
}
// If the sum of digits is
// not divisible by 3
if (sum % 3 != 0)
return false;
// Get the last digit of
// the number formed
int lastDigit = ((int)(str[n - 1]
- 'a' + 1))
% 10;
// If the last digit is
// not divisible by 2
if (lastDigit % 2 != 0)
return false;
return true;
}
// Driver code
int main()
{
string str = "ab";
int n = str.length();
if (isDivBySix(str, n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to return the sum
// of the digits of n
static int sumDigits(int n)
{
int sum = 0;
while (n > 0)
{
int digit = n % 10;
sum += digit;
n /= 10;
}
return sum;
}
// Function that return true if the
// decoded string is divisible by 6
static boolean isDivBySix(String str, int n)
{
// To store the sum of the digits
int sum = 0;
// For each character, get the
// sum of the digits
for (int i = 0; i < n; i++)
{
sum += (int)(str.charAt(i) - 'a' + 1);
}
// If the sum of digits is
// not divisible by 3
if (sum % 3 != 0)
return false;
// Get the last digit of
// the number formed
int lastDigit = ((int)(str.charAt(n - 1) -
'a' + 1)) % 10;
// If the last digit is
// not divisible by 2
if (lastDigit % 2 != 0)
return false;
return true;
}
// Driver code
public static void main(String []args)
{
String str = "ab";
int n = str.length();
if (isDivBySix(str, n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Rajput-Ji
Python 3
# Python3 implementation of the approach
# Function to return the sum
# of the digits of n
def sumDigits(n) :
sum = 0;
while (n > 0) :
digit = n % 10;
sum += digit;
n //= 10;
return sum;
# Function that return true if the
# decoded string is divisible by 6
def isDivBySix(string , n) :
# To store the sum of the digits
sum = 0;
# For each character, get the
# sum of the digits
for i in range(n) :
sum += (ord(string[i]) -
ord('a') + 1);
# If the sum of digits is
# not divisible by 3
if (sum % 3 != 0) :
return False;
# Get the last digit of
# the number formed
lastDigit = (ord(string[n - 1]) -
ord('a') + 1) % 10;
# If the last digit is
# not divisible by 2
if (lastDigit % 2 != 0) :
return False;
return True;
# Driver code
if __name__ == "__main__" :
string = "ab";
n = len(string);
if (isDivBySix(string, n)) :
print("Yes");
else :
print("No");
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the sum
// of the digits of n
static int sumDigits(int n)
{
int sum = 0;
while (n > 0)
{
int digit = n % 10;
sum += digit;
n /= 10;
}
return sum;
}
// Function that return true if the
// decoded string is divisible by 6
static bool isDivBySix(String str, int n)
{
// To store the sum of the digits
int sum = 0;
// For each character, get the
// sum of the digits
for (int i = 0; i < n; i++)
{
sum += (int)(str[i] - 'a' + 1);
}
// If the sum of digits is
// not divisible by 3
if (sum % 3 != 0)
return false;
// Get the last digit of
// the number formed
int lastDigit = ((int)(str[n - 1] -
'a' + 1)) % 10;
// If the last digit is
// not divisible by 2
if (lastDigit % 2 != 0)
return false;
return true;
}
// Driver code
public static void Main(String []args)
{
String str = "ab";
int n = str.Length;
if (isDivBySix(str, n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the sum
// of the digits of n
function sumDigits(n)
{
var sum = 0;
while (n > 0) {
var digit = n % 10;
sum += digit;
n = parseInt(n/10);
}
return sum;
}
// Function that return true if the
// decoded string is divisible by 6
function isDivBySix(str, n)
{
// To store the sum of the digits
var sum = 0;
// For each character, get the
// sum of the digits
for (var i = 0; i < n; i++) {
sum += (str[i].charCodeAt(0) - 'a'.charCodeAt(0) + 1);
}
// If the sum of digits is
// not divisible by 3
if (sum % 3 != 0)
return false;
// Get the last digit of
// the number formed
var lastDigit = ((str[n - 1].charCodeAt(0)
- 'a'.charCodeAt(0) + 1))
% 10;
// If the last digit is
// not divisible by 2
if (lastDigit % 2 != 0)
return false;
return true;
}
// Driver code
var str = "ab";
var n = str.length;
if (isDivBySix(str, n))
document.write( "Yes");
else
document.write( "No");
</script>
Output
Yes
时间复杂度 : O(N)
辅助空间: O(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
